Question

A solution containing $2.675g$ of $CoC{l}_3.6N{H}_3$ (Molar Mass = $267.5g/mol$) is passed through a cation exchanger. The chloride ions obtained in solution were treated with an excess of $AgN{O}_3$ to give $4.78g$ of $AgCl$ (Molar mass = $143.5g/mol$). The formula of the complex is

• A. $\left[CoC{l}_3\right(N{H}_3{)}_3]$
• B. $\left[CoCl\right(N{H}_3{)}_5]C{l}_2$
• C. $\left[Co\right(N{H}_3{)}_6]C{l}_3$
• D. $\left[CoC{l}_2\right(N{H}_3{)}_4]Cl$

Answer 1

The correct option is C $\left[Co\right(N{H}_3{)}_6]C{l}_3$

A cation exchanger exchanges the complex cation with another smaller cation. It is usually the size + charge that is used to do this.
Basically the nitrate of the silver nitrate is replaced by the chloride ions from the complex. You know the weights of the complex you start off with and the final weight of the silver chloride formed. Using simple mole concept here we can solve this.
So lets calculate the moles of chloride formed.
Moles of complex = $\frac{2.675}{267.5}=0.01$
Moles of $AgCl$ precipitated = $\frac{4.78}{143.5}=0.03$
$\therefore$ $C{l}^{-}$ ions produced = 0.03 moles
The above calculation shows that 3 moles of $C{l}^{-}$ ions are produced per mole of complex.

$\therefore$ Possible structure of complex is $\left[Co\right(N{H}_3{)}_6]C{l}_3$

$\left[Co\right(N{H}_3{)}_6]C{l}_3\to [Co(N{H}_3{)}_6{]}^{3+}+3C{l}^{-}$
1 Mole 3 Mole
0.01 mole 0.03 mole